您面临的挑战是打印输入，等待任何时间，打印输入，等待两次您最初等待的时间，再次打印输入，依此类推。初始延迟必须小于1小时，并且后续延迟的精度必须为+/- 5％。除此之外，对延迟时间没有限制。

例：

输入：hi。

输出：hi（1ms暂停）hi（2ms暂停）hi（4ms暂停）hi（8ms暂停）hi（16ms暂停），等等

还允许：

hi（暂停1分钟）hi（暂停2分钟）hi（暂停4分钟）hi（暂停8分钟）hi（暂停16分钟），等等。

输入必须在程序开始时提供（STDIN，命令行参数，函数参数等），并且必须是字符串。

初始延迟不能为0。

05AB1E，6个字节

码：

[=No.W

说明：

[        # Start an infinite loop
 =       # Print the top of the stack without popping
  No     # Compute 2 ** (iteration index)
    .W   # Wait that many milliseconds

在线尝试！

临时，8块+ 3字节

在Python中等效：

import time
n = 1
while 1:
    print("x")
    time.sleep(n)
    n = n * 2

Python 3中，60 56个字节

import time
def f(x,i=1):print(x);time.sleep(i);f(x,i*2)

变更日志：

• 将递归lambda更改为递归函数（-4字节）

MATL，8字节

`GD@WY.T

第一次暂停是2秒。

在MATL Online上尝试一下。或者查看修改后的版本，该版本显示自程序启动以来经过的时间。（如果解释器不起作用，请刷新页面，然后重试）。

或查看gif：

说明

`     % Do...while
  G   %   Push input
  D   %   Display
  @   %   Push iteration index (1-based)
  W   %   2 raised to that
  Y.  %   Pause for that time
  T   %   Push true. This will be used as loop confition
      % End (implicit). The top of the stack is true, which produces an infinite loop 

Mathematica 34 32 30 29字节

原始解决方案34字节：

For[x=.1,1<2,Pause[x*=2];Print@#]&

用Do删除2个字节

x=1;Do[Pause[x*=2];Print@#,∞]&

使用@MartinEnder的递归解决方案再减少一个字节

±n_:=#0[Print@n;Pause@#;2#]&@1

@ngenisis使用ReplaceRepeated递归删除另一个字节

1//.n_:>(Print@#;Pause@n;2n)&

八度，42 41字节

x=input('');p=1;while p*=2,pause(p),x,end

由于rahnema1，节省了一个字节，p*=2比还要短p=p*2。

我不敢相信我还不能打高尔夫球，但实际上并不是那么容易。

• 输入必须从头开始，因此第一部分不可避免。
• 我需要一个加倍的数字，并且必须在循环前对其进行初始化
  • 可以将输入用作循环的条件，但那时我将不得不 p*=2其他地方。
  • 暂停没有返回值，否则可能是 while pause(p*=2)

Java（OpenJDK 8），113字节

interface M{static void main(String[]a)throws Exception{for(int i=1;;Thread.sleep(i*=2))System.out.print(a[0]);}}

在线尝试！

-60个字节感谢Leaky Nun！

R, 50 48 bytes

function(x,i=1)repeat{cat(x);Sys.sleep(i);i=i*2}

returns an anonymous function which has one mandatory argument, the string to print. Prints no newlines, just spits x out on the screen. i is an optional argument that defaults to 1, waits for i seconds and doubles i.

-2 bytes thanks to pajonk

Try it online!

Ruby, 34 28 23 22 (+2 for -n) = 24 bytes

3 bytes saved thanks to Value Ink!

1 byte saved thanks to daniero

loop{print;sleep$.*=2}

Starts at 2, then 4, etc.

Explanation

-n                       # read a line from STDIN
  loop{                } # while(true):
       print;            # print that line
             sleep$.*=2  # multiply $. by 2, then sleep that many seconds. 
                         # $. is a Ruby special variable that starts at 1.

Alice, 16 bytes

1/?!\v
T\io/>2*.

Try it online! (Not much to see there of course, but you can check how often it was printed within one minute.)

Explanation

1    Push 1 to the stack. The initial pause duration in milliseconds.
/    Reflect to SE. Switch to Ordinal.
i    Read all input.
!    Store it on the tape.
/    Reflect to E. Switch to Cardinal.
>    Move east (does nothing but it's the entry of the main loop).
2*   Double the pause duration.
.    Duplicate it.
     The IP wraps around to the first column.
T    Sleep for that many milliseconds.
\    Reflect to NE. Switch to Ordinal.
?    Retrieve the input from the tape.
o    Print it.
\    Reflect to E. Switch to Cardinal.
v    Move south.
>    Move east. Run another iteration of the main loop.

R, 44 43 bytes

Crossed out 44 is still regular 44 ;(

This answer already provides a decent solution, but we can save some more bytes.

function(x)repeat{cat(x);Sys.sleep(T<-T*2)}

Anonymous function taking practically anything printable as argument x. Starts at 2 seconds and doubles every time afterwards. Abuses the fact that T is by default defined as TRUE which evaluates to 1.

Also, as long as this comment still gets a green light from OP, we can make it even shorter, but I don't think it is in the spirit of the challenge. Wait times of 0 are not allowed anymore.

function(x)repeat cat(x)

Cubix, 30 bytes

/(?:u<q.;1A>?ou2$/r;w;q^_q.\*/

Try it here

This maps onto a cube with side length 3.

      / ( ?              # The top face does the delay.  It takes the stack element with the
      : u <              # delay value, duplicates and decrements it to 0.  When 0 is hit the
      q . ;              # IP moves into the sequence which doubles the delay value.
1 A > ? o u 2 $ / r ; w  # Initiates the stack with one and the input.  For input hi this
; q ^ _ q . \ * / . . .  # gives us 1, -1, 10, 105, 104.  There is a little loop that prints 
. . . . . . . . . . . .  # each item in the stack dropping it to the bottom until -1 is hit.
      . . .              # Then the delay sequence is started om the top face
      . . .
      . . .

Bash, 37 bytes

for((t=1;;t*=2)){ sleep $t;echo $1;};

For some reason TIO won't show the output until you stop the program execution.

Try it online!

PHP, 31 bytes

for(;;sleep(2**$i++))echo$argn;
for(;;sleep(1<<$i++))echo$argn;

sleeps 1, 2, 4, 8, ... seconds. Run as pipe with php -nR '<code>'

Will work until the 63rd print (on a 64 bit machine), after that there will be no more waiting.
Version 1 will yield warnings sleep() expects parameter 1 to be integer, float given,
Version 2 will yield one warning sleep(): Number of seconds must be greater than or equal to 0.

Insert @ before sleep to mute the warnings.

TI-Basic, 21 bytes

Prompt Str0
1
While 1
Disp Str0
Wait Ans
2Ans
End

Python 3, 61 bytes

import time;i=1;x=input()
while 1:print(x);time.sleep(i);i*=2

Similar to @L3viathan's golf, but uses while loop

CJam, 26 bytes

qKes{es1$-Y$<{W$o;2*es}|}h

Doesn't work properly on TIO.

The first pause is 20 milliseconds.

Explanation

q                           e# Push the input.
 K                          e# Push 20 (the pause time).
  es                        e# Push the time (number of milliseconds since the Unix epoch).
    {                       e# Do:
     es1$-                  e#  Subtract the stored time from the current time.
          Y$<{              e#  If it's not less than the pause time:
              W$o           e#   Print the input.
                 ;2*es      e#   Delete the stored time, multiply the pause time by 2, push
                            e#     the new time.
                      }|    e#  (end if)
                        }h  e# While the top of stack (not popped) is truthy.
                            e#  (It always is since the time is a positive integer)

C, 51 bytes

main(c,v)char**v;{puts(v[1]);sleep(c);main(2*c,v);}

C, 35 bytes as a function

c=1;f(n){puts(n);sleep(c*=2);f(n);}

Takes input as a command line argument.

Batch, 62 bytes

@set/at=%2+0,t+=t+!t
@echo %1
@timeout/t>nul %t%
@%0 %1 %t%

This turned out to be a byte shorter than explicitly doubling t in a loop:

@set t=1
:g
@echo %1
@timeout/t>nul %t%
@set/at*=2
@goto g

Reticular, 12 bytes

1idp~dw2*~2j

Try it online!

Explanation

1idp~dw2*~2j
1               push 1 (initial delay)
 i              take line of input
  d             duplicate it
   p            print it
    ~           swap
     d          duplicate it
      w         wait (in seconds)
       2*       double it
         ~      swap
          2j    skip next two characters
1i              (skipped)
  d             duplicate input
   p            print...
                etc.

C#, 80 79 bytes

s=>{for(int i=1;;System.Threading.Thread.Sleep(i*=2))System.Console.Write(s);};

Saved one byte thanks to @raznagul.

Python 2, 54 bytes

Uses a lengthy calculation instead of timing libraries.

def f(x,a=1):
 while 1:a*=2;exec'v=9**9**6;'*a;print x

PowerShell, 35 33 30 29 Bytes

With a helpful hint from whatever and Joey

%{for($a=1){$_;sleep($a*=2)}}

Explanation

%{          # Foreach
for($a=1){  # empty for loop makes this infinite and sets $a
$_;         # prints current foreach item
sleep($a*=2)# Start-Sleep alias for $a seconds, reassign $a to itself times 2           
}}          # close while and foreach

Executed with:

"hi"|%{for($a=1){$_;sleep($a*=2)}}

Python 3, 49 bytes

b=input();x=6**6
while 1:print(b);exec("x+=1;"*x)

Uses the slight delay of the += operation and executes it x times. x doubles by adding one to itself as many times as the value of x.

It starts at 6^6 (46656) to stick to the maximum of 5% variation in the delay.

Perl 6, 39 bytes

print(once slurp),.&sleep for 1,2,4...* 

Try it (print overridden to add timing information)

Expanded:

  print(        # print to $*OUT
    once slurp  # slurp from $*IN, but only once
  ), 
  .&sleep       # then call sleep as if it was a method on $_

for             # do that for (sets $_ to each of the following)

  1, 2, 4 ... * # deductive sequence generator

JS (ES6), 44 42 40 38 36 bytes

^{Crossed out 44 is still 44}

i=1,y=x=>setTimeout(y,i*=2,alert(x))

Don't like alert bombs?

i=1,y=x=>setTimeout(y,i*=2,console.log(x))

Technically correct, but loophole-abusing:

y=x=>(x&&alert(x),y())

-3 bytes thanks to Cyoce, -2 thanks to Business Cat, -2 thanks to Neil

Common Lisp, 49 bytes

(do((a(read))(i 1(* 2 i)))(())(print a)(sleep i))

first delay should be 1 second.

Pyth, 7 bytes

#|.d~yT

Explanation:

#           Infinitely loop
  .d         Delay for 
      T      10 seconds
    ~y       and double T each time
 |           print input every iteration, too

TI-BASIC, 36 bytes

Initial wait period is 1 second.

1→L
Input Str1
checkTmr(0→T
While 1
While L>checkTmr(T
End
Disp Str1
2L→L
End

Racket, 51 bytes

(λ(s)(do([n 1(* n 2)])(#f)(displayln s)(sleep n)))

Example

➜  ~  racket -e '((λ(s)(do([n 1(* n 2)])(#f)(displayln s)(sleep n)))"Hello")'
Hello
Hello
Hello
Hello
Hello
^Cuser break